John, you said:

This is true for each prime $p$. But the integers, $\mathbf{Z},$ are more complicated than any of these $\mathbf{Z}/p$’s. To be precise, we have maps

$\mathbf{Z} \to \mathbf{Z}/p$

for each $p$. So, if we think of $\mathbf{Z}$ as a kind of space, it’s a big space that contains all the “planes” corresponding to the $\mathbf{Z}/p$’s. So, it’s 3-dimensional!

In short: from the viewpoint of ï¿½tale topology, the integers have one dimension that says which prime you’re at, and two more coming from the plane-like nature of each individual $\mathbf{Z}/p$.

I’m going to be a bit abrupt here and say that I think this point of view is wrong. Or at least it is so different to what I’m used to thinking that I would need much more detail to be convinced there is something to it.

Let me explain what I think is wrong and what I think the right way of thinking about it is.

First of all, you’ve used $\mathrm{Z}/p$ to mean two different things, and I think you’ve conflated them. There is $\mathrm{Spec}(\mathbf{Z}/p)$ and then there is the set of points of the affine line $\mathrm{Spec}(\mathbf{Z}/p[x])$ with coordinates in $\mathbf{Z}/p$. This is like confusing a point, whose function algebra is $\mathbf{C}$, with the space $\mathbf{C}$, whose function algebra is way bigger.

But you’re interested in $\mathrm{Spec}(\mathbf{Z})$. As you say, there are closed subspaces coming from the Spec$(\mathbf{Z}/p)$’s. Note that the affine line is nowhere in sight.

What about each $\mathrm{Spec}(\mathbf{Z}/p)$? They are each 1-dimensional from the point of view of the etale topology. Why? It is a basic fact, almost by definition, that a sheaf on $\mathrm{Spec}$ of a field is the same thing as a $G$-set, where $G$ is the absolute Galois group of the field. Therefore the etale cohomology is just the group cohomology of $G$. Therefore the etale cohomolical dimension of $\mathrm{Spec}$ of a field is the same thing as the cohomological dimension of its absolute Galois group. What is the abolute Galois group of a finite field? It’s the cyclic group freely generated by the Frobenius operator. And free cyclic groups have cohomological dimension 1. (I’m being a bit loose: there is more than one notion of cohomological dimension, and the absolute Galois group of a finite field is actually the free pro-cyclic group on one generator, but since we always look at cohomology with finite coefficients, we can ignore this.)

In other words, the $\mathbf{Z}/p$’s that cover $\mathbf{Z}$ are each one-dimensional, not two-dimensional.

So then why should there be the two dimensions of primes needed to make $\mathrm{Spec}(\mathbf{Z})$ three-dimensional? I don’t think there is a pure-thought answer to this question. As you wrote, there is a scientific answer in terms of Artin-Verdier duality, which is pretty much the same as class field theory. There is also a pure-thought answer to an analogous question. Let me try to explain that.

Instead of considering $\mathbf{Z}$, let’s consider $F[x]$, where $F$ is a finite field. They are both principal ideal domains with finite residue fields, and this makes them behave very similarly, even on a deep level. I’ll explain why $F[x]$ is three-dimensional, and then by analogy we can hope $\mathbf{Z}$ is, too. Now $F[x]$ is an $F$-algebra. In other words, $X=\mathrm{Spec}(F[x])$ is a space mapping to $S=\mathrm{Spec}(F)$. I already explained why $S$ is a circle from the point of view of the etale topology. So, if $X$ is supposed to be three-dimensional, the fibers of this map better be two-dimensional. What are the fibers of this map? Well, what are the points of $S$? A point in the etale topology is $\mathrm{Spec}$ of some field with a trivial absolute Galois group, or in other words, an algebraically closed field (even better, a separably closed one). Therefore a etale point of $S$ is the same thing as $\mathrm{Spec}$ of an algebraic closure $\bar{F}$ of $F$. What then is the fiber of $X$ over this point? It’s $\mathrm{Spec}$ of the ring $\bar{F}[x]$. Now, *this* is just the affine line over an algebraically closed field, so we can figure out its cohomological dimension. The affine line over the complex numbers, another algebraically closed field, is a plane and therefore has cohomological dimension 2. Since etale cohomology is kind of the same as usual singular cohomology, the etale cohomological dimension of Spec$(\bar{F}[x])$ ought to be 2.

Therefore $X$ looks like a 3-manifold fibered in 2-manifolds over Spec$(F)$, which looks like a circle. Back to Spec$(\mathbf{Z})$, we analogously expect it to look like a 3-manifold, but absent a (non-formal) theory of the field with one element, $\mathbf{Z}$ is not an algebra over anything. Therefore we expect Spec$(\mathbf{Z})$ to be a 3-manifold, but not fibered over anything.

## Re: This Week’s Finds in Mathematical Physics (Week 257)

Streater’s causes (lost and otherwise) can be found online here.